Calculate the antiderivative of a given functionConsider the function f : [ 0 , π 4 ) , f ( x ) = ( cos ⁡ x + sin ⁡ x ) n ( cos ⁡ x − sin ⁡ x ) n + 2 , where n ∈ N ∗ . Find the antiderivative F of f such that F ( 0 ) = 1 2 ( n + 1 ) .I've noticed that ( cos ⁡ x − sin ⁡ x ) ′ = − ( cos ⁡ x + sin ⁡ x ), but I can't figure out how to use this in finding F.

Question

Calculate the antiderivative of a given functionConsider the function   f  :      [    0    ,          π      4        )  ,   f  (  x  )  =            (      cos      ⁡      x      +      sin      ⁡      x              )        n                    (      cos      ⁡      x      −      sin      ⁡      x              )                  n          +          2                    , where   n  ∈            N        ∗  . Find the antiderivative F of f such that   F  (  0  )  =      1          2      (      n      +      1      )      .I&#039;ve noticed that   (  cos  ⁡  x  −  sin  ⁡  x      )    ′    =  −  (  cos  ⁡  x  +  sin  ⁡  x  ), but I can&#039;t figure out how to use this in finding F.

nuvolor8 · Accepted Answer

Step 1Noting   cos  ⁡  x  +  sin  ⁡  x  =      2    sin  ⁡  (  x  +      π    4    )  ,  cos  ⁡  x  −  sin  ⁡  x  =      2    cos  ⁡  (  x  +      π    4    )one has                     F        (        x        )                            =                    ∫                              (            cos            ⁡            x            +            sin            ⁡            x                          )              n                                            (            cos            ⁡            x            −            sin            ⁡            x                          )                              n                +                2                                                    d        x                                          =                    ∫                              (                          2                        sin            ⁡            (            x            +                          π              4                        )                          )              n                                            (                          2                        cos            ⁡            (            x            +                          π              4                        )                          )                              n                +                2                                                    d        x                                          =                    ∫                  tan          n                ⁡        (        x        +                  π          4                )                  1                      2                          cos              2                        ⁡            (            x            +                          π              4                        )                          d        x                                          =                              1          2                ∫                  tan          n                ⁡        (        x        +                  π          4                )                  sec          2                ⁡        (        x        +                  π          4                )        d        x                                          =                              1          2                ∫                  tan          n                ⁡        (        x        +                  π          4                )        d        tan        ⁡        (        x        +                  π          4                )                                          =                              1                      2            (            n            +            1            )                                    tan                      n            +            1                          ⁡        (        x        +                  π          4                )        +        C        .            Step 2Using   F  (  0  )  =      1          2      (      n      +      1      )      , it is easy to see   C  =  0.

Mayra Berry · Accepted Answer

Step 1We&#039;re trying to integrate  ∫  f  (  x  )  d  x  =  ∫            (      cos      ⁡      x      +      sin      ⁡      x              )        n                    (      cos      ⁡      x      −      sin      ⁡      x              )                  n          +          2                      d  x  ,,Divide numerator and denominator by       cos          n      +      2        ⁡  (  x  ):  ∫      1                  cos        2            ⁡      (      x      )                  (      1      +      tan      ⁡      x              )        n                    (      1      −      tan      ⁡      x              )                  n          +          2                      d  x  ,Now substitute   t  =  tan  ⁡  (  x  );   d  t  =      1                  cos        2            ⁡      (      x      )        d  x:                    ∫                              (            1            +            t                          )              n                                            (            1            −            t                          )                              n                +                2                                                    d        t                            =                  1          2                ∫                              (            1            +            t                          )              n                        (            1            −            t            +            1            +            t            )                                (            1            −            t                          )                              n                +                2                                                                                      =                  1          2                ∫                              (            1            +            t                          )              n                        (            1            −            t                          )              n                        (            (            1            −            t            )            +            (            1            +            t            )            )                                (            1            −            t                          )                              2                n                +                2                                                    d        t                                          =                  1          2                ∫                              (            1            −            t                          )                              n                +                1                                      (            1            +            t                          )              n                        +            (            1            +            t                          )                              n                +                1                                      (            1            −            t                          )              n                                            (            1            −            t                          )                              2                n                +                2                                                    d        t                                          =                  1                      2            (            n            +            1            )                          ∫                  d                      d            t                                                (                                          1                +                t                                            1                −                t                                      )                                n            +            1                          d        t                                          =                  1                      2            (            n            +            1            )                                                (                                          1                +                t                                            1                −                t                                      )                                n            +            1                          +        c                                          =                  1                      2            (            n            +            1            )                                                (                                          1                +                tan                ⁡                (                x                )                                            1                −                tan                ⁡                (                x                )                                      )                                n            +            1                          +        c            Since   tan  ⁡  (  0  )  =  0, you&#039;ll want   c  =  0.

Calculate the antiderivative of a given function Consider the function f : [ 0

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